How to use previous points calculated from a function.

Hello I have the following function

d=abs(B32 + A32*[X;Y]);sqrt(A32(1,1)^2+A32(1,2)^2);

the answer (d) changes based on the row matlab is on.

I have no problem calculating (d) based on the row.

What I want to do is use an equation R = d2*(t2-t1)/(d1-d2)

What I want to do is d2 to equal the current (d) that is being calculated.

I want d1 to equal the previous answer for (d).

I want the first d1 that is used to = 0 then the next d1 to equal (d) and so forth.

Is this possible.

Any help is greatly appreciated.

[Mitchell Thurston]

If I get the jist of what you're saying:

d1 = 0;
d2 = abs(B32 + A32*[X;Y]);
% it also looks like after the semicolon this is just being saved to "ans"

% .
% .
% .

% This is within some kind of loop
d2 = abs(B32 + A32*[X;Y]);
R = d2*(t2-t1)/(d1-d2);
d1 = d2;

Is this about what your asking for?

[Heraldo Tello]

when I'm calculating the R = d2*(t2-t1)/(d1-d2)

I want d2 = d = abs(B32 + A32*[X;Y]);sqrt(A32(1,1)^2+A32(1,2)^2);

another way to describe this is R = (current (d))*(t2-t1)/[(previous (d) that was calculated) - (current (d) that is calculated)]

d1 = the previous d that was calculated before the current (d)

however, when running this (loop I'm assuming), if d2 = the very first d that was calculated,

there is no previous (d) for d1 to take so I want d1 to initially be = 0 for this case. the next d1 will equal be the previous (d)

Answer 1

[Heraldo Tello]

% Plotting the TestPoints on the graph

for i = 1:size(TestPoints,1)

X=TestPoints(i,2);

Y=TestPoints(i,3);

Delta1=log(P1)-0.5*mu1*inv(Sw)*mu1'+mu1*inv(Sw)*[X;Y];

Delta2=log(P2)-0.5*mu2*inv(Sw)*mu2'+mu2*inv(Sw)*[X;Y];

Delta3=log(P3)-0.5*mu3*inv(Sw)*mu3'+mu3*inv(Sw)*[X;Y];

pause(1)

clc

% The first initial d1 = 0 because there is no previous distance. It's the

% first one. The next d1 will have the previous value of d2.

try

d1;

catch

d1=0;

end

% The first initial t1 = 0 because there is no previous time. It's the

% first one. The next t1 will have the previous value of t2.

try

t1;

catch

t1=0;

end

%

d2=abs(B32 + A32*[X;Y]);sqrt(A32(1,1)^2+A32(1,2)^2); % Distance to the LDA Boundary that separates aged and replace

t2=TestPoints(i,1); %time in seconds for the particular datapoint

%

%Calculating the Remaining Useful Life

RUL = d2*(t2-t1)/(d1-d2); % Units are in seconds

RUL_days=RUL/86400; % Units are in days

d1=d2;

t1=t2;

pause(1)

end

Answer 2

[KSSV]

You can solve the given equation:

R = d2*(t2-t1)/(d1-d2) ;

for d2 and solve it to get d2. We have d2 as:

d2 = R*d1/(R-t1+t2) ;

I assume that the (t2-t1) will be nothing but t(i+1)-t(i) .. if you use a loop; so this will be a time step and mostly will remain constant. Let it be dt. So the equation for d2 becomes:

d2 = R*d1/(R+dt) 

Now you need not use a loop..get the distance d1 and then find d2.

 d=abs(B32 + A32*[X;Y]);sqrt(A32(:,1).^2+A32(:,2).^2);   % 

The above formula should give you all d's at once. It might throw some error, in that case you need to give us more details about [X Y] and A32. Once d is obtained, use the equation for d2 and get d2.

[Heraldo Tello]

Sorry should have been more specific on the R formula. I am trying to solve for R. In order to do this I must know the current value of (d) and the past value of (d). I know what my t values are at each given point.